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3 years ago

Can somebody help and answer which ones are connected to each other. Correctly!

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

3 0

nuclear power--used to turn turbines...

fossil fuels--burned to provide energy that is....

renewable energy--energy that with come back after use

outlet--a device....

steam--nuclear reactors....

I'm not sure but I tried lol,lemme know if I'm wrong :D

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I really need help with this. Be serious. If you answer something random just for the points I will report you to brainly.

sashaice [31]

Answer:

It get thicker beacause the mid ocean

Explanation:

Because when it gets moved back the heat rises and it builds up to be thicker.

8 0

2 years ago

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.

adelina 88 [10]

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is $O_1H_1$

and written as OH.

8 0

3 years ago

Read 2 more answers

When an aqueous solution of silver nitrate is

Andre45 [30]

Answer:

$K^{+}$ and $NO_{3}^{-}$

Explanation:

The equation for the reaction is AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)

With all the ions, it is

$Ag^{+}$ (aq) + $NO_{3} ^{-}$ (aq) + $K^{+}$ (aq) + $Cl^{-}$ (aq) ==> AgCl(s) +

$K^{+}$ and $NO_{3} ^{-}$ do not change, so they are the spectator ions and are removed

The ionic equation is:

$Ag^{+}$ (aq) + $Cl^{-}$ (aq) ==> AgCl(s)

8 0

3 years ago

which group of the periodic table has elements with atoms that tend not to bond with other atoms of other elements

yuradex [85]

Noble gases:) they are very non-reactive

3 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago